Question

Question #37fe4

Answer 1

Here's what I would try.

Explanation:

Let's say that the sample that is too big to fit in the graduated cylinder has a volume `V` `"mL"`. After you weight the sample, you find that it has a mass `M` `"g"`.

By definition, the density of the mineral, `rho`, is equal to the ratio that exists between the mass of any sample of this mineral and the volume it occupies.

So for the initial sample, you have--I'll skip the units for simplicity

`rho = M/V" "" "color(darkorange)("(*)")`

Now, cut a piece of this sample that is small enough to fit in a graduated cylinder. Place this small piece on a scale and record its mass, let's say `m` `"g"`.

Use the graduated cylinder to determine its volume.

Let's say that this small piece has a volume `v` `"mL"`.

Since the density of the mineral must be the same regardless of the mass of the sample and the volume it occupies, you can say that--I'll skip the units for simplicity

`rho = m/v`

You can now use equation `color(darkorange)("(*)")` to say that

`M/V = m/v`

Rearrange to find `V`, the volume of the initial sample

`M/V = m/v implies V = (Mcolor(red)(cancel(color(black)("g"))))/(mcolor(red)(cancel(color(black)("g")))) * vcolor(white)(.)"mL"`

`V = (M/m * v)color(white)(.)"mL"`

And there you have it, the volume of the initial sample as a function of its mass `M` and the mass `m` and volume `v` of a smaller piece of this mineral.